Question: $\dfrac{ 4s + 2t }{ 9 } = \dfrac{ -4s - 3u }{ -2 }$ Solve for $s$.
Explanation: Multiply both sides by the left denominator. $\dfrac{ 4s + 2t }{ {9} } = \dfrac{ -4s - 3u }{ -2 }$ ${9} \cdot \dfrac{ 4s + 2t }{ {9} } = {9} \cdot \dfrac{ -4s - 3u }{ -2 }$ $4s + 2t = {9} \cdot \dfrac { -4s - 3u }{ -2 }$ Multiply both sides by the right denominator. $4s + 2t = 9 \cdot \dfrac{ -4s - 3u }{ -{2} }$ $-{2} \cdot \left( 4s + 2t \right) = -{2} \cdot 9 \cdot \dfrac{ -4s - 3u }{ -{2} }$ $-{2} \cdot \left( 4s + 2t \right) = 9 \cdot \left( -4s - 3u \right)$ Distribute both sides $-{2} \cdot \left( 4s + 2t \right) = {9} \cdot \left( -4s - 3u \right)$ $-{8}s - {4}t = -{36}s - {27}u$ Combine $s$ terms on the left. $-{8s} - 4t = -{36s} - 27u$ ${28s} - 4t = -27u$ Move the $t$ term to the right. $28s - {4t} = -27u$ $28s = -27u + {4t}$ Isolate $s$ by dividing both sides by its coefficient. ${28}s = -27u + 4t$ $s = \dfrac{ -27u + 4t }{ {28} }$